Question

If a particle is moving in a straight line such that its distance $s($in mts$)$ at any time $t($in sec) is given by $s=\frac{t^4}{4}-2t^3+4t^2+7,$ then its acceleration is minimum at which value of ${}^{\prime }{t}^{\prime }$

Answer 1

$s=\frac{t^4}{4}-2t^3+4t^2+7\Rightarrow v=\frac{ds}{dt}=t^3-6t^2+8t$
$\Rightarrow a=\frac{dv}{dt}=3t^2-12t+8$
$\frac{da}{dt}=6t-12$ and $\frac{d^2}{d{t}^2}=6>0$
${\therefore }^{\prime }{a}^{\prime }$ is minimum at $\frac{da}{dt}=0$
$\Rightarrow t=2$